The value of the integral ∫ _ 1 ^ 2 ( t ^ 4 + 1 t ^ 6 + 1 )dt is :

Given,

${∫}_{1}^{2}\dfrac{{t}^{4} + 1}{{t}^{6} + 1} d t$

$= {∫}_{1}^{2}\dfrac{{t}^{4} + 1}{\left({t}^{2} + 1\right)\left({t}^{4} - {t}^{2} + 1\right)} d t$

\left[∵ {a}^{3} + {b}^{3} = ( a + b ) ( {a}^{2} - ab + {b}^{2} )\right]$</p><p>$= {∫}_{1}^{2}\dfrac{{t}^{4} + 1 - {t}^{2} + {t}^{2}}{\left({t}^{2} + 1\right)\left({t}^{4} - {t}^{2} + 1\right)} d t$</p><p>$= {∫}_{1}^{2}\left[\dfrac{{t}^{4} + 1 - {t}^{2}}{\left({t}^{2} + 1\right)\left({t}^{4} - {t}^{2} + 1\right)} + \dfrac{{t}^{2}}{\left({t}^{2} + 1\right)\left({t}^{4} - {t}^{2} + 1\right)}\right] d t$</p><p>$= {∫}_{1}^{2}\left[\dfrac{1}{\left({t}^{2} + 1\right)} + \dfrac{{t}^{2}}{\left({t}^{2} + 1\right)\left({t}^{4} - {t}^{2} + 1\right)}\right] d t$</p><p>$= {∫}_{1}^{2}\left[\dfrac{1}{\left({t}^{2} + 1\right)} + \dfrac{{t}^{2}}{{t}^{6} + 1}\right] d t$</p><p>$= {∫}_{1}^{2}\dfrac{1}{\left({t}^{2} + 1\right)}dt + {∫}_{1}^{2}\dfrac{{t}^{2}}{{\left({t}^{3}\right)}^{2} + 1} d t$</p><p>$= {\left[\tan^{- 1}t\right]}_{1}^{2} + \dfrac{1}{3}{∫}_{1}^{2}\dfrac{3{t}^{2}}{{\left({t}^{3}\right)}^{2} + 1} d t$</p><p>$= \left[\tan^{- 1}2 - \tan^{- 1}1\right] + \dfrac{1}{3}{\left[\tan^{- 1}{t}^{3}\right]}_{1}^{2}$</p><p>$= \left[\tan^{- 1}2 - \dfrac{\pi}{4}\right] + \dfrac{1}{3}\left[\tan^{- 1}8 - \tan^{- 1}1\right]$</p><p>$= \left[\tan^{- 1}2 - \dfrac{\pi}{4}\right] + \dfrac{1}{3}\left[\tan^{- 1}8 - \dfrac{\pi}{4}\right]$</p><p>$= \tan^{- 1}2 - \dfrac{\pi}{4} + \dfrac{\tan^{- 1}8}{3} - \dfrac{\pi}{12}$</p><p>$= \tan^{- 1}2 + \dfrac{\tan^{- 1}8}{3} - \dfrac{\pi}{3}$